Convergence/divergence of $\sum_{k=1}^\infty\frac{2\times 4\times 6\times\cdots\times(2k)}{1\times 3\times 5\times\cdots\times(2k-1)}$ A problem asks me to determine if the series
$$\sum_{k=1}^\infty \frac{2 \times 4 \times 6 \times \cdots \times (2k)}{1 \times 3 \times 5 \times \cdots \times (2k-1)}$$
converges or diverges.
(from the textbook Calculus by Laura Taalman and Peter Kohn (2014 edition); section 7.6, p. 639, problem 33)
I am allowed to use the ratio test first and then any other convergence/divergence test if the former test does not work. In my original work, I attempted the ratio test and it was rendered inconclusive.
$$ a_k = \frac{2 \times 4 \times 6 \times \cdots \times (2k)}{1 \times 3 \times 5 \times \cdots \times (2k-1)} $$
$$ a_{k + 1} = \frac{2 \times 4 \times 6 \times \cdots \times (2k) \times (2(k+1))}{1 \times 3 \times 5 \times \cdots \times (2k-1) \times (2(k+1)-1)} = \frac{2 \times 4 \times 6 \times \cdots \times (2k) \times (2k+2)}{1 \times 3 \times 5 \times \cdots \times (2k-1) \times (2k+1)} $$
$$ \frac{a_{k + 1}}{a_k} = \frac{\frac{2 \times 4 \times 6 \times \cdots \times (2k) \times (2k+2)}{1 \times 3 \times 5 \times \cdots \times (2k-1) \times (2k+1)}}{\frac{2 \times 4 \times 6 \times \cdots \times (2k)}{1 \times 3 \times 5 \times \cdots \times (2k-1)}} = \frac{2 \times 4 \times 6 \times \cdots \times (2k) \times (2k+2)}{1 \times 3 \times 5 \times \cdots \times (2k-1) \times (2k+1)} \times \frac{1 \times 3 \times 5 \times \cdots \times (2k-1)}{2 \times 4 \times 6 \times \cdots \times (2k)} = \frac{2k+2}{2k+1}$$
Evaluating $\rho = \lim_{x \to \infty} \frac{a_{k + 1}}{a_k}$ will determine if $\sum_{k=1}^\infty \frac{2 \times 4 \times 6 \times \cdots \times (2k)}{1 \times 3 \times 5 \times \cdots \times (2k-1)}$ converges or diverges. The conclusions for the ratio test are as follows:
$\circ$ If $\rho < 1$, then $\sum_{k=1}^\infty a_k$ converges.
$\circ$ If $\rho > 1$, then $\sum_{k=1}^\infty a_k$ diverges.
$\circ$ If $\rho = 1$, then the test is inconclusive.
$$ \rho = \lim_{x \to \infty} \frac{a_{k + 1}}{a_k} = \lim_{x \to \infty} \frac{2k+2}{2k+1} = 1$$
Since $\rho = 1$, the ratio test is rendered inconclusive, as I stated earlier. I will have to use other convergence/divergence tests to solve the problem.
My issue is that I'm not sure which other convergence/divergence test to use. Any suggestions?
Many thanks for the help.
 A: The series is divergent since $a_k>1$ for every $k$.
A: By Stirling approximation, we have that
$$a_k=\frac{2\times 4\times 6\times\cdots\times(2k)}{1\times 3\times 5\times\cdots\times(2k-1)}=\frac{4^k\cdot (k!)^2}{(2k)!}=\frac{4^k}{\binom{2k}{k}}\sim \sqrt{\pi k}.$$
Hence the series $\sum_{k\geq 1} a_k$ diverges.
More generally the series $\sum_{k\geq 1} \frac{a_k}{k^r}$ converges iff $r-1/2>1$, that is $r>3/2$.
A: $$\sum_{k=1}^\infty \frac{2 \times 4 \times 6 \times \cdots \times (2k)}{1 \times 3 \times 5 \times \cdots \times (2k-1)}=$$
$$\sum_{k=1}^\infty \frac{\prod_{j=1}^k(2j)}{\prod_{j=1}^k(2j-1)}=$$
$$\sum_{k=1}^\infty \prod_{j=1}^k\frac{(2j)}{(2j-1)}$$
The product $a_k = \prod_{j=1}^k\frac{(2j)}{(2j-1)}$ is
$$\left(1+\dfrac11\right)\left(1+\dfrac13\right)\left(1+\dfrac15\right)\cdots \left(1+\dfrac1{2n-1}\right)$$
An infinite product $\lim_{k \to \infty} \left(1+a_k\right)$ converges to a non-zero number if one of the $\sum_{k \to \infty} \vert a_k \vert$ converges. 
Conclude what you want from this.
